The question is below?

# mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is the distance from the plane to the top of larger sphere?

1 Answer
Jun 28, 2018

The distance is 2\sqrt(2) + 322+3 or \approx 5.8284271255.828427125

Explanation:

Let the radiuses of the mutually tangent spheres be A(1,1)A(1,1) and B(3,1)B(3,1)
respectively

Let C(2,y)C(2,y) be the radius of the sphere that lies on top of them

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overline{AC} = 3 ¯¯¯¯¯¯AC=3 and overline{BC} = 3 ¯¯¯¯¯¯BC=3
Recall the distance formula:
d = sqrt{ (x_1 - x_2)^2 + (y_1-y_2)^2 } d=(x1x2)2+(y1y2)2

Thus,
1^2 + (y-1)^2 = 3^2 12+(y1)2=32
1 + (y-1)^2 = 9 1+(y1)2=9
(y-1)^2 = 8 (y1)2=8
y = 1 \pm 2sqrt{2} y=1±22

Since y y must be greater than 1 because it is on top of the spheres, reject the negative solution
y = 2sqrt{2} + 1y=22+1

Since the radius of the sphere is 2, h = y+2 h=y+2
height h = y + 2 = 2sqrt{2} + 1 + 2 h=y+2=22+1+2

\therefore The distance is 2\sqrt(2) + 3